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Old 02-03-2013, 11:23 PM   #1391 (permalink)
flyweightnate
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Join Date: Oct 2007
Location: Dallas

Quote:
Originally Posted by P0E View Post
You guys might be over thinking the max efficiency numbers. Classical physics should be adequate considering all work in should come out. Conservation of energy and all that stuff. If you want to consider the thermodynamic process then you can look at both isothermal and adiabatic by just adjusting your assumptions.

Lets start with the the PV of free air and consider the compression process first. If compression was completely an adiabatic process (because PV remains perfectly insulated) all work performed on PV simply increases heat content of the gas itself. This additional heat increases the temperature/pressure. PV after compression is therefore much higher than PV before. Note this means the pressure of a 20,400ci volume of free air compressed to 68ci would be much greater than 4500psi.

Isothermal assumes the tank is not perfectly insulated so all of this heat energy flows into the atmosphere. As the tank sits there losing heat it is going from an adiabatic process to an isothermal one. PV before compression equals PV after compression once all heat exits the tank and it reaches ambient temperature. In the above example, pressure at this point WOULD now equal 4500psi.

Both isothermal and adiabatic processes come up with the same amount of potential maximum energy when used correctly. The difference is in the location of this stored energy. Isothermal stores it in the atmosphere and adiabatic stores it in the tank. Regardless, think of the gas as an energy converter instead of an energy storage device.

You can tell to what degree the compression process was isothermal or adiabatic based on how hot your tank is. We know how much work we did and the degree to which that energy would increase the gas temperature. Since most players consider a 'full' tank one that has reached the maximum fill pressure at ambient temperature, I'll say compression should be thought of as entirely isothermal and therefore all the energy is actually stored external to the tank in the form of heat in the atmosphere.

Regardless of the process, isothermal or adiabatic process, all work done to the system was converted and stored as heat about to be recovered. The compression is perfectly reversible since we live in a perfect world while calculating theoretical limits. So all we need to do is use some classical physics to determine how much work done to, or could be done by, a cylinder piston using force and distance. Once again, this is the boiled down formula:

E = P1*V*ln(P1/P2)

Now lets look at when the gas is put to work.

The adiabatic equations should work when calculating the energy inside of a tank about to undergo an adiabatic process, assuming you use the appropriately large temperatures and pressures. Remember after the tank was compressed, all heat stayed inside the tank.

So what if we assumed compression was isothermal and expansion is adiabatic? Well then energy in the tank would be the same before compression as after, just like the isothermal process. We're just ignoring the fact that the atmosphere has energy which our gas can convert to work because we are now assuming our tank is perfectly insulated so no heat will flow in.

Although the isothermal expansion process implies gas is released slow enough that heat flows into the tank at the same rate work work is performed... I'll bet we can tell to what degree that's accurate. i.e. What ratio the system is isothermal versus adiabatic. How many more shots do you get when you fire your gun once per day? If the increase is negligible, your fire rate was already mostly isothermal.

I realize this post has already turned into a novel, but let me continue onto the CO2.

I started with the mass of CO2 then calculated it's volume at ambient temp and pressure. I then calculated the amount of work performed to compress it just prior to the phase change. The phase change would be additional potential energy in the form of P deltaV which could be added as minor increase.

Now that should be a sufficient explanation as to why the following numbers are valid:

Energy in a 68ci 4500psi HPA tank: ~198KJ
Roughly 16k shots max depending on ball mass and velocity.

Energy in a 12g CO2 cartridge: ~2500J
Roughly 200 shots max depending on ball mass and velocity.


Note that if you want to take it a step past the tank, inside the gun.... i.e. argue that because the gas is released in 5mS it MUST be adiabatic because no heat can be transferred to the CO2 that quickly.... well then you'll have to provide some evidence to that argument.
Funny, I did this math and got about 2800 for a 68/45k, and 65 for a 12g. 300fps, 3.2g paintball? Drag? Linear acceleration, viscous drag... ? There's a reason nobody's gone past 3k with a 68/45k- and if theoretical was 16,000 shots per tank, I'm sure I could do at least 10,000. I already built the gun. And didn't get 1000 shots on my 13/3k.
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